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Thread: [SOLVED] Integral formulas

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Integral formulas

    Hi MHF!
    I need your help for some questions.
    First, what is the method to find out that $\displaystyle \int \frac{dx}{1+x^2}=arctan(x)+C$ ? I find it stupid to learn it by heart. There are plenty of other formulas to know, so I would like to know how to recover them without learning all of them.
    Second, I have to find $\displaystyle \int \frac{dx}{\sqrt{x^2-9}}$ in term of trigonometric or hyperbolic functions. Looking at my book, I found out that it is equal to $\displaystyle ln|x+\sqrt{x^2-9}|+C$. Which recalls me of an inverse of $\displaystyle cosh^{-1}$ but cannot find exactly to what it's equal, so I'd appreciate help for this one.
    And finally $\displaystyle \int \frac{dx}{\sqrt{4-x^2}}=arcsin\big(\frac{x}{2}\big)+C$ according to my book. But once again, how can I know it's worth an inverse of the sine function? I hope I don't have to learn this by heart.
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  2. #2
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    Here is a way to look at it and probably the method commonly used.

    $\displaystyle \frac{d}{dx}[tan^{-1}(x)]=\frac{1}{1+x^{2}}$

    $\displaystyle y=tan^{-1}(x), \;\ x=tan(y)$

    Differentiating the right one implicitly gives us:

    $\displaystyle \frac{d}{dx}[x]=\frac{d}{dx}[tan(y)]$

    $\displaystyle 1=sec^{2}(y)\cdot\frac{dy}{dx}$

    $\displaystyle \frac{dy}{dx}=\frac{1}{sec^{2}(y)}=\frac{1}{sec^{2 }(tan^{1}(x))}.....[1]$

    Using the triangle we see that:

    $\displaystyle sec^{2}(tan^{-1}(x))=1+x^{2}$

    And [1] simplifies to $\displaystyle \frac{dy}{dx}=\frac{1}{1+x^{2}}$

    Therefore, $\displaystyle \int\frac{1}{1+x^{2}}=tan^{-1}(x)$

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    To find $\displaystyle \int\frac{1}{\sqrt{x^{2}-9}}dx$

    let $\displaystyle x=3sec{\theta}, \;\ dx=3sec{\theta}tan{\theta}d{\theta}$

    When you make the subs and simplify it whittles down to

    $\displaystyle \int sec{\theta}d{\theta}$

    Then, to get back in terms of x just resub $\displaystyle {\theta}=sec^{-1}(\frac{x}{3})$

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~

    $\displaystyle \int\frac{1}{\sqrt{4-x^{2}}}dx$

    This one simplifies wonderfully. Let $\displaystyle x=2sin{\theta}, \;\ dx=2cos{\theta}d{\theta}$

    Make the subs and you finally get:

    $\displaystyle \int d{\theta}={\theta}$

    What is theta?. $\displaystyle {\theta}=sin^{-1}(x/2)$

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~

    To tell when to use what, just remember the formulas:

    $\displaystyle \sqrt{a^{2}-x^{2}}$, then use $\displaystyle x=a\cdot sin{\theta}$

    $\displaystyle \sqrt{a^{2}+x^{2}}$, then use $\displaystyle x=a\cdot tan{\theta}$

    $\displaystyle \sqrt{x^{2}-a^{2}}$, then use $\displaystyle x=a\cdot sec{\theta}$

    Can you see whey we use those when we do?. Because they result in the proper identities that get rid of the radical.
    Last edited by galactus; Nov 24th 2008 at 05:38 AM.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks a lot, I'll be analyzing this closely.
    You made an error of typo in the line "$\displaystyle \frac{dy}{dx}=\frac{1}{sec^{2}(y)}=\frac{1}{sec^{2 }(tan^{1}(x))}$.....[1]" Must be $\displaystyle tan^{-1}$. Otherwise I think all is O.K. Thanks so much!!!
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  4. #4
    MHF Contributor Reckoner's Avatar
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    Smile

    Quote Originally Posted by arbolis View Post
    And finally $\displaystyle \int \frac{dx}{\sqrt{4-x^2}}=arcsin\big(\frac{x}{2}\big)+C$ according to my book. But once again, how can I know it's worth an inverse of the sine function? I hope I don't have to learn this by heart.
    Let $\displaystyle y = \arcsin x$. Then we have

    $\displaystyle x = \sin y$

    $\displaystyle \Rightarrow \cos y\frac{dy}{dx} = 1$

    $\displaystyle \Rightarrow \frac{dy}{dx} = \frac1{\cos y} = \frac1{\sqrt{1 - \sin^2 y}} = \frac1{\sqrt{1 - x^2}}$

    The other inverse functions are similar.
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  5. #5
    Senior Member nikhil's Avatar
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    Basic thinking

    hi!arbolis,hope ur having a great time.
    Hmmmmm..... You have asked a very basic question.though i can tell you how to integrate some function by a definite procedure (like for your first integral question put x=tany,for second put X=3secy etc) but i think i should give you basics that are behind the procedures
    The main problem is that unlike differentiation there is no definite first principle for integration.when we perform integration,we take it as a reverse process of differentiation. Like
    d(sinx)/dx=cosx
    therefor
    integral cosxdx=sinx.
    So as you can see if you know the derivative of a function then the integral of its derivative is the function itself.so whenever we need to integrate a complex looking function we try to make it less complex by substitution.
    Like A= $\displaystyle \int \frac{dx}{1+x^2}$
    let x=tany
    differentiating both sides we get
    dx=(secy)^2
    now substituting these value we get
    A=$\displaystyle
    \int\frac{(secy)^{2} dy}{1+(tany)^2}
    $
    but 1+(tany)^2=(secy)^2
    so A= $\displaystyle
    \int dy
    $ [so because of the substitution integral became easy]
    so A=y+c
    since x=tany
    y=arctanx
    so A=arctanx+c
    i hope this will help in widening your thinking.
    REMEMBER it may not be possible to solve all questions by substitution.so like differentiation there are some other procedures[like partial fraction method}.
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  6. #6
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    It is really helpful to memorize some basic forms. For example the derivative of $\displaystyle \sin^{-1}$ (on $\displaystyle (-1,1)$). The way we do this is to noting that $\displaystyle \sin (\sin^{-1} x ) = x$ thus $\displaystyle (\sin^{-1} x) \cos (\sin^{-1} x) = 1$ by the chain rule. But, $\displaystyle \cos (\sin^{-1} x) = \sqrt{1-x^2}$ which means $\displaystyle (\sin^{-1} x) ' = \tfrac{1}{\sqrt{1-x^2}}$. Here is an example, $\displaystyle \smallint \tfrac{dx}{\sqrt{9-x^2}}$. We can factor this, $\displaystyle \tfrac{1}{3}\smallint \tfrac{dx}{\sqrt{1-(\frac{x}{3})^2}}$. Now let $\displaystyle t=\tfrac{x}{3}$ and we get $\displaystyle \smallint \tfrac{dt}{\sqrt{1-t^2}} = \sin^{-1} t + k = \sin^{-1} \tfrac{x}{3}+k$.

    Now find a similar form for $\displaystyle \sinh^{-1} x$. Again, $\displaystyle \sinh (\sinh^{-1} x) = x$ by the chain rule we get $\displaystyle (\sinh^{-1} x)' \cosh(\sinh^{-1} x) = 1$. Now note that $\displaystyle \cosh(\sinh^{-1} x) = \sqrt{x^2+1}$*. For example, $\displaystyle \smallint \tfrac{dx}{\sqrt{x^2+4}}$ can be solved in a similar way as above problem. Inside of doing all those substitutions.


    *)Let $\displaystyle f(x) = \cosh(\sinh^{-1} x)$ and see that $\displaystyle f(x)>0$. But then $\displaystyle f^2(x) = \cosh^2 (\sinh^{-1} x) = 1 + \sinh^2(\sinh^{-1} x) = 1+x^2$. Thus, $\displaystyle f(x) = \sqrt{x^2+1}$.
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